An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 115 cm and a standard deviation of 4.7 cm. A. Find the probability that one selected subcomponent is longer than 117 cm. Probability = B. Find the probability that if 4 subcomponents are randomly selected, their mean length exceeds 117 cm. Probability = C. Find the probability that if 4 are randomly selected, all 4 have lengths that exceed 117 cm.

a)

And we can find this probability using the complement rule and we got:

b)

And using the complement rule and a calculator, excel or the normal standard table we have that:

c)

And the reason is because we assume independence between the events and each individual event have the same probability of success 0.197.

And we want this probability:

We can use the probability mass function and we got:

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

And the best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

And we can find this probability using the complement rule and we got:

Part b

For this case we select a sample size of n=4, since the distribution of X is normal then the distribution for is also normal and given by:

We can find the individual probabilities like this:

And using the complement rule and a calculator, excel or the normal standard table we have that:

Part c

For this case we can use a binomial distribution given by:

And the reason is because we assume independence between the events and each individual event have the same probability of success 0.197.

And we want this probability:

We can use the probability mass function and we got: